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The Ka for acid HA is 7.0 × 10⁻⁴. what is the pOH of a 0.10 M aqueous solution of Ka?

a) 5.92
b) 1.00
c) 8.08
d) 1.58
e) 5.01
f) 11.92
g) 2.08

User Cupakob
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1 Answer

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Final answer:

To determine the pOH of a 0.10 M solution of acid HA with a Ka of 7.0 × 10⁻⁴, one would perform equilibrium calculations using the provided Ka to find [H+] and subsequently pH, before using the relationship pH + pOH = 14 to find the pOH.

Step-by-step explanation:

The question asks for the pOH of a 0.10 M aqueous solution of a weak acid HA with a known Ka value of 7.0 × 10⁻⁴. To find this, one would use the acid dissociation constant (Ka) to determine the concentration of hydroxide ions ([OH-]). Given the relationship of pH + pOH = 14, once the pH is found, pOH can be easily calculated.

Because HA is a weak acid, it partially disassociates in water to form H+ and A-. If we assume that the concentration of H+ is equal to x, then the equilibrium concentrations can be set up as [HA] = 0.10-x, [H+] = x, and [A-] = x. The Ka expression would be Ka = [H+][A-]/[HA], and we would solve for x to determine [H+], and consequently the pH.

However, without going through the calculations, it is not possible to determine the exact pOH from the given options. Therefore, a full calculation must be done considering all equilibrium expressions and constants to arrive at the correct answer.

User Esac
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